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Let $X/\mathbb{Q}$ be an irreducible smooth projective curve with a $\mathbb{Q}$-rational point $p$. Then there is a map $\phi: X \rightarrow \textrm{Pic}^{0}(X)$ with the property that $q \mapsto [q]-[p]$. If I recall correctly this map induces an isomorphism on Galois representations $H^1_{et}(X\times \bar{\mathbb{Q}}, \mathbb{Q}_\ell) \rightarrow H^1_{et}(\textrm{Pic}^{0}(X)\times \bar{\mathbb{Q}}, \mathbb{Q}_\ell)$ (for any $\ell$ where $X$ has good reduction).

I am pretty interested in working through a careful proof of this (likely well known) fact but but I don't know how I would check something like it. I am hoping someone can give me pointers/references so I can do that.

There are many natural generalizations one could expect. Can you give references (maybe in SGA 4 or Milne's Etale Cohomology or BLR) so I can check for myself what the most general version of this is (eg, can one somehow allow some $\ell$ of bad reduction, take maps from surfaces to their albanese, or etale cohomology of $X$ instead of $X\times \bar{\mathbb{Q}}$.)

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    $\begingroup$ This is true over any separably closed field with any $\ell$. The key is that it is true with coefficients in any finite abelian group: every abelian connected finite Galois cover of $X$ arises by $\phi$-pullback from a unique such cover of the Jacobian $J$. Milne discusses this with references in his article on Jacobians in the book "Arithmetic Geometry". Or use deformation of curves and smooth/proper base change to reduce to char. 0, and then to $\mathbf{C}$ where it reduces to determining the analytification of $(J, \phi)$ in terms of that of $X$. $\endgroup$
    – user76758
    Commented Dec 5, 2013 at 5:58
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    $\begingroup$ Of course, to argue by passage to char. 0 you have to assume $\ell$ is not the characteristic (so restrict to abelian connected Galois covers with degree not divisible by the characteristic). $\endgroup$
    – user76758
    Commented Dec 5, 2013 at 6:20

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